question_answer

A circular loop wire with current \[{{i}_{1}}=({{i}_{0}}/\pi )\] and a V shaped wire with current;, are arranged in a plane as shown in diagram. If magnetic field at O is zero, value of \[{{\operatorname{i}}_{2}}\] is:

A) \[\frac{13a{{i}_{0}}}{35r}\]

B) \[\frac{12a{{i}_{0}}}{17r}\]

C) \[\frac{a{{i}_{0}}}{r}\]

D) \[\frac{12a{{i}_{0}}}{35r}\]

Correct Answer: D

Solution :

here, \[a=r'\left( \cot 37{}^\circ +\cot 53{}^\circ  \right)\]

\[={{r}^{'}}\left( \frac{4}{3}+\frac{3}{4} \right)\]

\[\therefore {{r}^{'}}=\frac{12a}{25}\]

Magnetic field due to-V-shaped conducter

\[{{B}_{1}}=\left[ \frac{{{\mu }_{0}}}{4\pi }.\frac{{{i}_{2}}}{\frac{12a}{25}}\left( \sin 53{}^\circ +\sin 37{}^\circ  \right) \right]\]\[=2\left[ \frac{{{\mu }_{0}}}{4\pi }.\frac{{{i}_{2}}}{\frac{12a}{25}}\times \frac{7}{5} \right]\] or \[70\left( \frac{{{\mu }_{0}}}{4\pi } \right)\frac{{{i}_{2}}}{12a}\]

According to given condition

\[{{B}_{1}}={{B}_{2}}\]

\[70\left( \frac{{{\mu }_{0}}}{4\pi } \right)\frac{{{i}_{2}}}{12a}=\frac{{{\mu }_{0}}\left( \frac{{{i}_{0}}}{\pi } \right)}{2r}\therefore {{i}_{2}}=\frac{12a{{i}_{0}}}{35r}.\]